# Inital-value problem

1. Mar 11, 2008

### morbello

Have been looking at this problem .A few times nowI've come up with different answers.Its getting close to needing to know or wether, to just put in what i have.so ive placed a question on here to see if im missing something.

The equations that are said to help are

$$\int\f'(x)/f(x) dx = In (f(x))+c (f(x)>0)$$

dy$$/dx = x^4+1/x^5+5x+6 (x>-1)$$ is the equation to work off.

my work so far is

1$$/5 In (x^5+5)+c$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 11, 2008

### rock.freak667

3. Mar 12, 2008

### morbello

so the x^5+5x+6 does not differeniate to become x^5+5 ok

there are a few activitys in my books that are like it.

4. Mar 12, 2008

### rock.freak667

$$\int \frac{x^4+1}{x^5+5x+6} dx$$

Let $t=x^5+5x+6 \Rightarrow \frac{dt}{dx}=5(x^4+1)$

$\frac{dt}{5}=(x^4+1)dx$

$$\int \frac {x^4+1}{x^5+5x+6} dx \equiv \int \frac{1}{5} \frac{1}{t} dt$$

and $\int \frac{1}{x} dx = ln(x)+C$

$$\int \frac{1}{5} \frac{1}{t} dt = \frac{1}{5} lnt + C$$

5. Mar 12, 2008

### morbello

that blew me a little bit there.ive sat and looked at it a little and i think i know what you have done just i would not have thought off it. i think we did one function that was worked out that way.ill have to look it up tonight.